Please correct me if I am wrong, but I really don't see where we need the proper assumption to get a Galois action.
So let $X/k$ be any scheme, $\overline{k}$ a separable (or algebraic) closure of $k$, and $\mathcal{F}$ be any sheaf on $X$. Write $X_\overline{k}$ for the base change of $X$, $p:X_\overline{k}\rightarrow X$ the induced morphism. Write also $\mathcal{F}_\overline{k}=p^*\mathcal{F}$ for the pullback of $\mathcal{F}$ to $X_\overline{k}$.
For $g\in\mathrm{Gal}(\overline{k}/k)$, we have an induce morphism $g:X_\overline{k}\rightarrow X_\overline{k}$, which induce a pullback $g^*\mathcal{F}_\overline{k}$. I claim that there is a canonical isomorphism $g^*\mathcal{F}_\overline{k}\simeq\mathcal{F}_\overline{k}$. This is because $\mathcal{F}$ comes from $X$. Indeed : $g^*\mathcal{F}_\overline{k}=g^*p^*\mathcal{F}=p^*\mathcal{F}=\mathcal{F}_\overline{k}$.
Hence we have an induced action on cohomology :
$$ H^i(X_\overline{k},\mathcal{F}_\overline{k})\rightarrow H^i(X_\overline{k},g^*\mathcal{F}_\overline{k})\simeq H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
which is the action you are looking for.
Now there is also the approach of Alex Youcis : if $f:X\rightarrow\operatorname{Spec}k$ is the structural morphism, $R^if_*\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$ hence a set equipped with a continuous discrete $\operatorname{Gal}(\overline{k}/k)$-set. Unless I'm mistaken, I don't think we need to add any assumption for the following claim : the underlying $\operatorname{Gal}(\overline{k}/k)$-set is exactly $H^i(X_\overline{k},\mathcal{F}_\overline{k})$ with the above action. (This prove in particular that the above action is continuous).
This follows from the following continuity result
$$\varinjlim_{k'}H^i(X_{k'},\mathcal{F}_{k'})=H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
where $k'$ runs through the finite Galois extensions of $k$ in $\overline{k}$ and the trivial base change $(R^if_*\mathcal{F})_{k'}=R^if_*\mathcal{F}_{k'}$.
EDIT : Let me expand a bit on the last assertion. First, let us show that the stalk of $R^nf_*\mathcal{F}$ at a geometric point $\operatorname{Spec}\overline{k}$ is indeed $H^n(X_\overline{k},\mathcal{F}_\overline{k})$.
By definition, the stalk of $R^nf_*\mathcal{F}$ is $\varinjlim_{k'}R^nf_*\mathcal{F}(k')$ where the limit is taken over all the finite extension of $k'$ inside $\overline{k}$. Now recall that $R^nf_*\mathcal{F}$ is the sheaf associated to the presheaf $k'\mapsto H^n(X_{k'},\mathcal{F}_{k'})$, and since the stalk of a presheaf is the same as its associated sheaf, we get $(R^nf_*\mathcal{F})_\overline{k}=\varinjlim_{k'}H^n(X_{k'},\mathcal{F}_{k'})=H^n(X_\overline{k},\mathcal{F}_\overline{k})$, the last equality is from a limit argument (we should add $X$ quasi-compact and quasi-separated here).
Recall that the equivalence between sheaves on $(\operatorname{Spec}k)_{ét}$ and $\operatorname{Gal}(\overline{k}/k)$-sets is the following : if $\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$, then $\mathcal{F}_\overline{k}=\varinjlim\mathcal{F}(k')$ is a $\operatorname{Gal}(\overline{k}/k)$-set. The action of $\sigma\in\operatorname{Gal}(\overline{k}/k)=\varprojlim\operatorname{Gal}(k'/k)$ on $\mathcal{F}_\overline{k}$ is induced by the compatibles actions on the $\mathcal{F}(k')$ (check that is indeed compatible).
Now you should convinced yourself that this is indeed the pull-back to $\overline{k}$ and that this action is the same as the induced action by functoriality (that is very first one I wrote). Do the same with $R^if_*$ and you will get the compatibility of the two actions in general.
Exercise : Take $k=\mathbb{R}$ and $\mathcal{F}=\mu_4$ the sheaf $A\mapsto\{x\in A, x^4=1\}$ and compute the action of the stalk (there is no trap here).
Best Answer
as already mentioned, if $X$ is defined over the ground field $K$, the Galois group acts trivially on these equations. Maybe its nice to think of a few examples:
Take a finite Galois extension $L/K$ of deg d, say there is a $K$ basis $\{\lambda_1,\dots, \lambda_d\}$ Then define the hypersurface S in $\mathbf{A}_K^d$ by the product equation $$\prod_{\sigma \in Gal(L/K)}(\lambda_1^\sigma x_1+\dots+\lambda_d^\sigma x_d)$$ Geometrically, this is a union $d$ hyperplanes conjugate via the Galois action and you see that $G$ acts on the coefficients of the equations of$S$,i.e. $$\lambda^\sigma := \sigma(\lambda)$$ But: $S$ does not have any $K$ points, because any solution $\alpha \in K$ would imply $\lambda_1^\sigma\alpha_1 +\dots \lambda_n^\sigma\alpha_n=0$ for $\alpha_i \in K$ which is impossible because the $\lambda_i$ form a $K$ basis of $L$.
Take the affine line $\mathbf{A}^1$ over $\mathbf{F}_p$, then $\sigma=\mathrm{Frob}_p \in \mathrm{Gal}_{\mathbf{F}_p}$ acting on e.g.$(t-\alpha) \mapsto (t-\alpha^p)$ for $\alpha \in \bar{\mathbf{F}_p}$. Then this action is trivial if $\alpha \in \mathbf{F}_p$. So for the fixed set under this action $\mathbf{A}^1(\bar{\mathbf{F}}_p){^\sigma}=\mathbf{A}^1(\mathbf{F}_p)$. Note that this Frobenius element is a topological generator of the absolute Galois group as a profinite group so it tells you a lot about it.
This is finally an example of an étale sheaf and a Galois action on it: Namely, fix a field $k$ with algebraic closure $K$ and let $G$ be the absolute Galois group of $k$. Let $\mathcal{F}$ be an étale sheaf (more precisely, a sheaf in the finite étale topology) on $X:=\textrm{Spec}(k)$ and define $\mathcal{F}_K := \varinjlim\mathcal{F}(k')$ where the limit runs over all finite separable extensions $k'$ of $k$. Then $\mathcal{F}_K$ carries a natural $G$ action such that for each section $f \in \mathcal{F}_K$ the subgroup $\{\sigma \in G: f^\sigma = f\}$ is of finite index so this $G$ action is called continuous inducing an equivalence of categories $$ \{\text{sheaves on} X_{fét}\} \rightarrow \{\text{sets with cont. G-action}\} $$ coming from sending $\mathcal{F} \mapsto \mathcal{F}_K$ . To see how this action works, just observe that on each $\mathcal{F}(k')$, $\sigma$ acts after restriction to $k'$ by pullback $\mathcal{F}(k') \mapsto \sigma|_{k'}^*\mathcal{F}(k')$. Note that the subgroup $\textrm{Gal}(K|k')$ of $G$ acts trivially on $\mathcal{F}_k'$ having finite index #$\textrm{Gal}(k'|k)$. This example is from Olssons book on algebraic spaces p.66